The Weierstrass Approximation Theorem - College of Engineering ...

MATH 4540: Analysis Two

The Weierstrass Approximation Theorem

James K. Peterson

Department of Biological Sciences and Department of Mathematical Sciences Clemson University

February 26, 2018

MATH 4540: Analysis Two

Outline

1 The Wierstrass Approximation Theorem 2 MatLab Implementation 3 Compositions of Riemann Integrable Functions

MATH 4540: Analysis Two The Wierstrass Approximation Theorem

The next result is indispensable in modern analysis. Fundamentally, it states that a continuous real-valued function defined on a compact set can be uniformly approximated by a smooth function. This is used throughout analysis to prove results about various functions. We can often verify a property of a continuous function, f , by proving an analogous property of a smooth function that is uniformly close to f . We will only prove the result for a closed finite interval in . The general result for a compact subset of a more general set called a Topological Space is a modification of this proof which is actually not that more difficult, but that is another story.

MATH 4540: Analysis Two The Wierstrass Approximation Theorem

Theorem

Let f be a continuous real-valued function defined on [0, 1]. For any > 0, there is a polynomial, p, such that |f (t) - p(t)| < for all t [0, 1], that is || p - f ||<

Proof

We first derive some equalities. We will denote the interval [0, 1] by I . By the binomial theorem, for any x I , we have

n n xk (1 - x)n-k = (x + 1 - x)n = 1.

()

k

k =0

MATH 4540: Analysis Two The Wierstrass Approximation Theorem

Proof Differentiating both sides of Equation , we get

nn 0=

k

k =0

kx k-1(1 - x )n-k - x k (n - k)(1 - x )n-k-1

= n n x k-1(1 - x )n-k-1 k(1 - x ) - x (n - k) k

k =0

= n n x k-1(1 - x )n-k-1 k - nx k

k =0

Now, multiply through by x(1 - x), to find

0 = n n xk (1 - x)n-k (k - nx). k

k =0

Differentiating again, we obtain

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